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Chapter 7: Problem 13

The greatest value of \(f(x)=\cos \left(x e^{|x|}+7 x^{2}-3 x\right)\), $x \in[-1, \infty)$ is (A) \(-1\) (B) 1 (C) 0 (D) None of these

Short Answer

Expert verified
Answer: (D) None of these.

Step by step solution

01

Recall properties of the cosine function

The cosine function returns values in the interval [-1, 1], with the greatest value being 1. This occurs when the argument inside the cosine function is a multiple of 2蟺 (i.e., \(\cos(2n\pi) = 1\) for n being an integer).
02

Analyze the given function

We have the function \(f(x) = \cos(x e^{|x|} + 7x^{2} - 3x)\). To find the greatest value of this function over the interval [-1, 鈭�), we will try to find when the argument inside the cosine function is equal to \(2n\pi\), for n being an integer.
03

Determine when the argument is a multiple of \(2\pi\)

We need to determine when \(x e^{|x|} + 7x^{2} - 3x = 2n\pi\) for some integer n. However, this is not a straightforward equation to solve analytically.
04

Try an approximated argument

Instead, let's try a closer interval for x and check if it's possible for the expression \(x e^{|x|} + 7x^{2} - 3x\) to be equal to \(2n\pi\): For \(x\in[-1, 1]\): - \(x e^{|x|}\) ranges between 0 and 1, the entire function's variation is small. - \(7x^{2}\) ranges between 0 and 7, which is insufficient for a cosine function that has period \(2\pi \approx 6.28\) - \(3x\) ranges between -3 and 3, not enough as well. This exploratory process shows that within this close interval, it's highly unlikely that we will obtain an argument of the cosine function that is a multiple of \(2\pi\).
05

Conclude the answer

Based on the steps above, we cannot find an x value in the range [-1, 鈭�) such that the argument inside the cosine function is a multiple of 2蟺, which means the function \(f(x) = \cos(x e^{|x|} + 7x^{2} - 3x)\) cannot achieve its potential maximum value of 1. Therefore, the greatest value of the function given the domain is: (D) None of these

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Most popular questions from this chapter

Let $f(x)=\left\\{\begin{array}{lr}x^{3}+x^{2}-10 x, & -1 \leq x<0 \\ \cos x, & 0 \leq x<\pi / 2 \\ 1+\sin x, & \pi / 2 \leq x \leq \pi\end{array}\right.$ Then \(\mathrm{f}(\mathrm{x})\) has (A) a local minimum at \(x=\pi / 2\) (B) a global maximum at \(\mathrm{x}=-1\) (C) an absolute minimum at \(\mathrm{x}=-1\) (D) an absolute maximum at \(\mathrm{x}=\pi\)

Consider $\mathrm{f}(\mathrm{x})=|1-\mathrm{x}| 1<\mathrm{x}<2 \quad 1 \leq \mathrm{x} \leq 2$ and \(g(x)=f(x)+b \sin \pi / 2 x, \quad 1 \leq x \leq 2\) then which of the following is correct? (A) Rolles Theorem is applicable to both \(\mathrm{f}, \mathrm{g}\) and \(\mathrm{b}=3 / 2\) (B) LMVT is not applicable to \(\mathrm{f}\) and Rolle's Theorem if applicable to \(g\) with \(b=1 / 2\) (C) LMVT is applicable to \(\mathrm{f}\) and Rolle's Theorem is applicable to \(g\) with \(b=1\) (D) Rolle's Theorem is not applicable to both \(\mathrm{f}, \mathrm{g}\) for any real \(\underline{b}\)

$f(x)=\left\\{\begin{array}{cc}2-\left|x^{2}+5 x+6\right|, & x \neq-2 \\\ a^{2}+1 & , x=-2\end{array}\right.$, then the range of a so that \(f(x)\) has maxima at \(x=-2\) is \((\) A) \(|a| \geq 1\) (B) \(\mid \mathrm{a}<1\) (C) \(\mathrm{a}>1\) (D) \(\mathrm{a}<1\)

A cylindrical gas container is closed at the top and open at the botton. If the iron plate of the top is \(5 / 4\) times as thick as the plate forming the cylindrical sides, the ratio of the radius to the height of the cylinder using minimum material for the same capacity is (A) \(3 ; 4\) (B) \(5 ; 6\) (C) \(4: 5\) (D) None of these

Let \(f(x)=a x^{3}+b x^{2}+c x+d(a, b, c, d, \alpha, \beta \in R)\) and $\mathrm{f}^{\prime}(\alpha)=\mathrm{f}^{\prime}(\beta)=0,(\alpha \neq \beta)$, then (A) if \(f(\alpha), f(\beta)<0\) then all the roots of \(f(x)=0\) are real (B) if \(f(\alpha), f(\beta)>0\) then all the roots of \(f(x)=0\) is real (C) if \(\alpha\) is a point of local maxima and \(\beta\) is point of local minima for the function \(f(x)\) then \(\alpha<\beta\) (D) If \(a>0\) then \(f(x)\) is decreasing on \((\alpha, \beta)\), where \(\beta>\alpha\)
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